1
$\begingroup$

I recently knew of this note in which Prof. M. Henk presents a proof of Minkowski's second inequality on successive minima which is (purportedly) based on ideas in Minkowski's original proof. Let me remind you that, for a centrally symmetric convex body $\mathcal{K} \subseteq \mathbb{R}^{n}$ and a lattice $\Lambda \subseteq \mathbb{R}^{n}$, the said inequality asserts that

$$(\lambda_{1} \cdots \lambda_{n}) \mathrm{vol}(\mathcal{K}) \leq 2^{n}\mathrm{det}(\Lambda)$$

where $\lambda_{i}$ is the $i$-th successive minimum of the body $\mathcal{K}$ with respect to the lattice $\Lambda$.

At the outset of the proof, Prof. Henk makes $K_{i} := \frac{\lambda_{i}}{2}\mathcal{K}$ for every $i \in \{1, \ldots,n \}$ and picks up $z^{1}, \ldots, z^{n} \in \mathbb{Z}^{n}$ in such a way that $z^{1} \in \lambda_{1}\mathcal{K}, \ldots, z^{n} \in \lambda_{n}\mathcal{K}$ and the set $\{z^{1}, \ldots, z^{n}\}$ is linearly independent. Moreover, he notes that the $z^{i}$ can be chosen so as to satisfy $$\mathbf{span}(z^{1}, \ldots, z^{i})=\mathbf{span}(e_{1}, \ldots, e_{i})=:L_{i}$$ where $\{e_{1}, \ldots, e_{n}\}$ is the standard basis of $\mathbb{R}^{n}$. Then, after defining $M_{q}^{n}$ as the set $\{u \in \mathbb{Z}^{n} \colon |u_{i}|\leq q, \, 1\leq i \leq n\}$ and $M_{q}^{i}$ as the intersection $M_{q}^{n} \cap L_{i}$, it is effortlessly established that $$\mathrm{vol}(M_{q}^{n}+K_{n}) \leq (2q+\gamma)^{n}$$ (for some positive constant $\gamma$) and that $$\mathrm{vol}(M_{q}^{n}+K_{1}) =(2q+1)^{n}\mathrm{vol}(K_{1}).$$

Then, it comes the important observation according to which the desired inequality would easily follow from what has just been proven and from the inequalities $$\mathrm{vol}(M_{q}^{n}+K_{i+1}) \geq \left(\frac{\lambda_{i+1}}{\lambda_{i}}\right)^{n-i} \mathrm{vol}(M_{q}^{n}+K_{i}) \qquad \qquad ... (*)$$

In my opinion, it is the verification of the inequalities in $(*)$ one of the key steps in Prof. Henk's proof of Minkowski's second inequality on successive minima. Unfortunately, even when it may be possible to validate line-by-line the proof which he provides for them, I must confess that it is not clear to me what the overall idea behind it is... In case you grasped it through and through once, would you be so kind as to motivate and/or explain below, in plain English, the underlying idea (or ideas) beneath Henk's proof of the inequalities in $(*)$?

Please, let me thank you in advance for your insightful replies, pointers to the literature, etc.

$\endgroup$
2
$\begingroup$

As far as I can see it, there are two ideas behind $(\star)$:

First, one argues that it's enough to consider the volumes for $M_q^i$ instead of $M_q^n$. This is the case, because everything is defined in such a way that $K_j$ won't overlap with any translate by a $(\mathbb{Z}^n\cap\langle e_{i+1},...,e_n\rangle)$-vector, if $j\geq i$.

Thus, one can think of the packing $M_q^n+K_{i+1}$ as interiorly disjoint layers of $M_q^i+K_{i+1}$ piled up over one another, and the same for $M_q^n+K_{i}$, where the number of layers is in both cases $(2q+1)^{n-i}$. So it is enough to consider those $i$-dimensional layers.

Next, one essentially uses the fact that $K_{i+1}$ equals $K_i$ scaled by $\lambda_{i+1}/\lambda_{i}$. But instead of scaling all the coordinates 'at the same time', we may only sacale the first $i$, place the partly scaled body on the $M_q^i$-points and then scale the remaining $n-i$ coordinates. This won't affect the packing set $M^i_q$, since it sits in the first $i$ coordinates. This 'scaling-after-packing' causes the factor $(\frac{\lambda_{i+1}}{\lambda_i})^{n-i}$.

('Packing' is probably not the best expression here, since the translates overlap...)

To make it very short and plain, the (rough) plan for proving $(\star)$ is:

Restrict to $i$-dimensional layers of the packing and then scale $K_i$ up to $K_{i+1}$ in a 'careful way'.

I hope this helps a bit.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.